Consider 4 positions of four-digit PIN:
1. PIN cannot begin with 0, then it can begin with 9 remaining digits (1, 2, 3, 4, 5, 6, 7, 8 and 9) - 9 posibilities;
2. for the second and third position yuo can select any of 10 digits (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) - 10·10=100 posibilities;
3. the PIN must be an even number and cannot be 0, then it ends with any of 4 even digits (2, 4, 6 and 8) - 4 posibilities.
Use the product rule to count the total number of posibilities:
9·100·4=3600.
Answer: correct choice is A.
